PDF of $\frac{\min(X,Y)}{\max(X,Y)}$ when $X$ and $Y$ are iid Exponential with parameter $\lambda$

Question

I am given that $X$ and $Y$ are iid Exponential with parameter $\lambda$. Then I am to compute pdf of $\frac{\min(X,Y)}{\max(X,Y)}$

I have derived that the pdf of $Z=\min(X,Y)$ . is given by:-

$$f_{Z}(z)=2\lambda e^{-2\lambda z}\,\, , z\geq 0.$$

And $T=\max(X,Y)$ is given by

$$f_{T}(t)=2\lambda (1-e^{-\lambda t})e^{-\lambda t}\,\,, t\geq 0.$$

Now to derive the PDF of $\frac{Z}{T}$ I am thinking of using the Jacobian method with a substitution of $U$ and $V$ such that $U=\frac{Z}{T}$ and $V=Z$ . and then integrating the joint pdf of $U,V$ over $v$ to get the pdf of $U$. But I cannot find the joint pdf of $Z,T$ if Z and T are not independent.

Is my thinking correct?. Am I in the right direction?. What is the correct method to solve this ?.

Answer 1

But I cannot find the joint pdf of Z,T if Z and T are not independent.

The joint pdf is just:

$$\begin{align}f_{Z,T}(z,t) &=f_{\min(X,Y),\max(X,Y)}(z,t) \\[1ex] &= f_X(Z)f_Y(T)+f_X(T)f_Y(Z) \\[1ex] &= 2\lambda^2\mathrm e^{-\lambda(z+t)}\mathbf 1_{0\leq z\leq t}\end{align}$$

---

So the marginal of $Z\div T$ can be found by :

$$\begin{align}f_{Z\div T}(s) &= \mathbf 1_{0\leq s\leq 1}\int_0^\infty \left\lvert\tfrac{\partial st}{\partial s}\right\rvert f_{Z,T}(st,t)\,\mathrm d t\\[1ex]&=2\lambda^2\,\mathbf 1_{0\leq s\leq 1}\int_0^\infty t~\mathrm e^{-\lambda(1+s)t}\,\mathrm d t \\[1ex]&= \phantom{2(1+s)^{-2}\,\mathbf 1_{0\leq s\leq 1}} \end{align}$$

Answer 2

The random variable $W:=\frac{\min(X,Y)}{\max(X,Y)}$ has support in $[0,1]$. Let $0\le t\le1$. Then \begin{align*} P(W\le t)&=P(\min(X,Y)\le t\max(X,Y))\\[.4em]&=P(X\le tY,X<Y)+P(Y\le tX,X>Y)\\[.4em] &=P(X\le tY)+P(Y\le tX)\\[.4em] &=E[1-\mathrm e^{-\lambda tY}]+E[1-\mathrm e^{-\lambda tX}]\\[.4em] &=2-2E[\mathrm e^{-\lambda tX}]\\[.4em] &=2-\frac2{1+t}, \end{align*} so a density is $$f_W(t):=\frac{\mathrm d}{\mathrm dt}P(W\le t)=\frac2{(1+t)^2},\quad0\le t\le1.$$